Answer Happy

Carefully go through the steps of the proof below. Then answer the questions at the end. Claim. Let x be real number with x € [0, 1]. Then sin x + cos x ≥ 1. Proof. ; by (1) u € [0,7] u arbitrary 9 (2) (3) (4) (5) (6) (7) 1+2 sin u cos u < 1 (8) 2 sin u cos u < 0 (9) either sin u <0 or cos u < 0 sin u ≥ 0 and cos u ≥ 0, for u € [0, 7] sin u + cos u < 1 (sin u + cos u)² < 1² sin² u + 2 sin u cos u + cos² u < 1 (sin² u + cos² u) + 2 sin u cos u < 1 π (10) u = [0, (11) Vr[x = [0,] ⇒ sinx+cos x ≥ 1] ⇒sin u + cos u ≥ 1 ; by ; by ; by ; by ; by ; by ; by ; by ; by ; by_ (1) Decide whether this is: (i) direct proof (ii) proof by contraposition – (iii) proof by contradiction - (iv) proof by cases.
(2) Decide whether the proof is correct or not. (2a) If the proof is incorrect, identify the first line where a mistake happens. (2b) If the proof is correct, fill in the blanks by choosing from the following list: (i) by algebra – (ii) by assumption - (iii) by closure – (iv) by definition - (v) by definition, for some suitable choice of k (when you enter your answers on Brightspace, you may use the abbreviated form in boldface) – (vi) by logic. [2P]